Question

Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 

Answer 1

\[\text{Let I }= \int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]         ...........(1)

Then,

\[I = \int_0^{2\pi} \log\left[ \sec\left( 2\pi - x \right) + \tan\left( 2\pi - x \right) \right]dx ...............\left[ \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[ = \int_0^{2\pi} \log\left( \sec x - \tan x \right)dx .....................\left( 2 \right)\]

Adding (1) and (2), we get

\[2I = \int_0^{2\pi} \left[ \log\left( \sec x + \tan x \right) + \log\left( \sec x - \tan x \right) \right]dx\]
\[ \Rightarrow 2I = \int_0^{2\pi} \log\left( \sec^2 x - \tan^2 x \right)dx\]
\[ \Rightarrow 2I = \int_0^{2\pi} \log1dx .................\left( 1 + \tan^2 x = \sec^2 x \right)\]
\[ \Rightarrow 2I = 0 ......................\left( \log1 = 0 \right)\]
\[ \Rightarrow I = 0\]